Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $y = \dfrac{-3a - 12}{a + 1} \times \dfrac{-2a - 18}{a^2 + 13a + 36} $
Solution: First factor the quadratic. $y = \dfrac{-3a - 12}{a + 1} \times \dfrac{-2a - 18}{(a + 4)(a + 9)} $ Then factor out any other terms. $y = \dfrac{-3(a + 4)}{a + 1} \times \dfrac{-2(a + 9)}{(a + 4)(a + 9)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac{ -3(a + 4) \times -2(a + 9) } { (a + 1) \times (a + 4)(a + 9) } $ $y = \dfrac{ 6(a + 4)(a + 9)}{ (a + 1)(a + 4)(a + 9)} $ Notice that $(a + 9)$ and $(a + 4)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac{ 6\cancel{(a + 4)}(a + 9)}{ (a + 1)\cancel{(a + 4)}(a + 9)} $ We are dividing by $a + 4$ , so $a + 4 \neq 0$ Therefore, $a \neq -4$ $y = \dfrac{ 6\cancel{(a + 4)}\cancel{(a + 9)}}{ (a + 1)\cancel{(a + 4)}\cancel{(a + 9)}} $ We are dividing by $a + 9$ , so $a + 9 \neq 0$ Therefore, $a \neq -9$ $y = \dfrac{6}{a + 1} ; \space a \neq -4 ; \space a \neq -9 $